Question

If $z$ is a complex number, then the number of solution(s) for the equation $z^2=\overline{z}$ is

• A. $1$
• B. $2$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

Let $z=x+iy$
As $z^2=\overline{z}$
$\Rightarrow (x+iy{)}^2=x-iy\Rightarrow x^2-y^2+2ixy=x-iy$
Equality of complex numbers holds when
$x^2-y^2=x,2xy=-y\Rightarrow 2xy=-y\Rightarrow y(2x+1)=0\Rightarrow y=0,x=-\frac{1}{2}$

When $y=0$, then
$x^2-y^2=x\Rightarrow x(x-1)=0\Rightarrow x=0,1$

When $x=-\frac{1}{2}$, then
$x^2-y^2=x\Rightarrow \frac{1}{4}-y^2=-\frac{1}{2}\Rightarrow y^2=\frac{3}{4}\Rightarrow y=\pm \frac{\sqrt{3}}{2}$

Therefore,
$(x,y)=(0,0),(1,0),(-\frac{1}{2},\frac{\sqrt{3}}{2}),(-\frac{1}{2},-\frac{\sqrt{3}}{2})$

Hence, there $4$ complex numbers satisfying the given equation.